Linear inverse problems and theirregularisation
Umut Çetin
London School of Economics

Bachelier Workshop, September 2016

Umut Çetin

Linear inverse problems and their regularisation

An inverse problem for Markov processes
Suppose X is an E-valued Markov process withsemi-group (Pt )t≥0 and infinitesimal generator A, where E
is some topological space. Given a function g and T > 0
does there exists an f so thatg(x) = PT f (x) := E x [f (XT )]?

Umut Çetin

(1)

Linear inverse problems and their regularisation

An inverse problem for Markov processes
Suppose X is an E-valued Markov process withsemi-group (Pt )t≥0 and infinitesimal generator A, where E
is some topological space. Given a function g and T > 0
does there exists an f so thatg(x) = PT f (x) := E x [f (XT )]?

(1)

If the answer to the above is affirmative, then we candefine a function u : [0, T ] × E 7→ R by u(t, ·) = PT −t f which
will solve
ut + Au = 0;u(0, ·) = g;(2)

a backward PDE with an initial condition!
Umut Çetin

Linear inverse problems and their regularisation

Backward equations with an initial condition
A class of ill-posed PDEs akin to the one given in (2) has
been recently studied in Mathematical Finance literature in
the context of forward utilities (see, e.g., the works by
Nadtochiy as well as Tehranchi and Shkolnikov et al.) .
However, these works search for solutions of (2) over
[0, ∞) while the linear inverse problem (1) is related to theL2 -solutions over a bounded interval [0, T ].

Umut Çetin

Linear inverse problems and their regularisation

Backward equations with an initial condition
A class of ill-posed PDEs akin to the one given in (2) has
been recently studied in Mathematical Finance literature in
the context of forward utilities (see, e.g., the works by
Nadtochiy as well as Tehranchi and Shkolnikov et al.) .
However, these works search for solutions of (2) over
[0, ∞) while the linear inverse problem (1) is related to theL2 -solutions over a bounded interval [0, T ].
This is a crucial distinction as a solution over [0, ∞)necessarily require the initial condition, g, to be unbounded
while the inverse problem (1) with a solution in L2 implicitly
imposes boundary conditions on g given by the behaviourof the diffusion. In particular, g must vanish at the naturalor absorbing boundaries.

Umut Çetin

Linear inverse problems and their regularisation

If g is a density then the linear inverse problem withpositivity constraint is the answer to the following question:
Given the European option price data at time-T can wefind an initial distribution for our favourite diffusion process
to obtain a model consistent with the observed option
prices?
As such, the solutions of (1) provide alternative “answers"
to certain questions in Mathematical Finance that are
traditionally attacked using Skorokhod embedding.

Umut Çetin

Linear inverse problems and their regularisation

If g is a density then the linear inverse problem withpositivity constraint is the answer to the following question:
Given the European option price data at time-T can wefind an initial distribution for our favourite diffusion process
to obtain a model consistent with the observed option
prices?
As such, the solutions of (1) provide alternative “answers"
to certain questions in Mathematical Finance that are
traditionally attacked using Skorokhod embedding.
CAUTION: If (1) has a solution for a positive g, the solution
is not necessarily positive:
Consider X being Brownian motion and g(x) = x 2 .
Then, f (x) = x 2 − T .

Umut Çetin

Linear inverse problems and their regularisation

One dimensional diffusions
In order to make the discussion precise let’s suppose X isa one-dimensional diffusion on a natural scale on (l, r ) withd2
a given boundary behaviour, i.e. A = 12 σ 2 dx
2 where σ is anonnegative measurable function. The associated speed
measure, m, on (l, r ) has density 2/σ 2 (x)dx on the interior,
I, of the domain.

Umut Çetin

Linear inverse problems and their regularisation

One dimensional diffusions
In order to make the discussion precise let’s suppose X isa one-dimensional diffusion on a natural scale on (l, r ) withd2
a given boundary behaviour, i.e. A = 12 σ 2 dx
2 where σ is anonnegative measurable function. The associated speed
measure, m, on (l, r ) has density 2/σ 2 (x)dx on the interior,
I, of the domain.
Then, it is well-known that A is symmetric with respect tom.
That is, if f and h are in D(A) ∩ L2 (I, m) ∩ S, where S is theclass of functions satisfying the boundary conditions
stipulated by the behaviour of the diffusion at the boundary,then(Af , h) = (f , Ah),R

where (f , h) := I f (x)h(x)m(dx) is the inner product on theHilbert space H = L2 (I, m).
Umut Çetin

Linear inverse problems and their regularisation

The spectrum of A
Consider the eigenvalue problem
Aφ = −λφ,φ ∈ S.

The values of λ for which the above equation has anontrivial solution make up the spectrum of A.
If none of the boundaries are natural it was shown by
Joanne Elliott (1954) that the spectrum of A is discrete.
Moreover, there exists a function p(t, x, y ) symmetric in x
and y and continuous in all variables such that for any
continuous f vanishing at the boundaries of I we haveZ

Pt f (x) =Umut Çetin

p(t, x, y )m(dy ).
I

Linear inverse problems and their regularisation

The function p(t, x, y ) has the representationp(t, x, y ) =∞X

e−λn t φn (x)φn (y ),n=0where ) ≤ λ0 ≤ λ1 ≤ . . . λn ≤ . . . are the eigenvalues
increasing to ∞ and φn s are corresponding eigenfunctions.
Moreover, (φn ) is an orthonormal sequence, i.e kφn k = 1
and (φn , φm ) = 0 for n 6= m.

Umut Çetin

Linear inverse problems and their regularisation

The function p(t, x, y ) has the representationp(t, x, y ) =∞X

e−λn t φn (x)φn (y ),n=0where ) ≤ λ0 ≤ λ1 ≤ . . . λn ≤ . . . are the eigenvalues
increasing to ∞ and φn s are corresponding eigenfunctions.
Moreover, (φn ) is an orthonormal sequence, i.e kφn k = 1
and (φn , φm ) = 0 for n 6= m.
The spectral expansion of the transition density inpresence of natural boundaries is obtained by H. P.
McKean. In this case the spectrum is not necessarily
discrete and the eigenfunctions are not necessarily square
integrable.

Umut Çetin

Linear inverse problems and their regularisation

Introducing the inverse operator
To make Rthe presentation simpler let’s redefine Pt so thatPt f (x) = I p(t, x, y )f (y )m(dy ). (Pt ) is still a kernel but it is
sub-Markovian, i.e. Pt 1 < 1 if one of the boundaries areabsorbing.

Umut Çetin

Linear inverse problems and their regularisation

Introducing the inverse operator
To make Rthe presentation simpler let’s redefine Pt so thatPt f (x) = I p(t, x, y )f (y )m(dy ). (Pt ) is still a kernel but it is
sub-Markovian, i.e. Pt 1 < 1 if one of the boundaries areabsorbing.
To ease the exposition we will assume that there are nonatural boundaries but all the results will continue to holdwhen a natural boundary is present after appropriate
modifications.

Umut Çetin

Linear inverse problems and their regularisation

Introducing the inverse operator
To make Rthe presentation simpler let’s redefine Pt so thatPt f (x) = I p(t, x, y )f (y )m(dy ). (Pt ) is still a kernel but it is
sub-Markovian, i.e. Pt 1 < 1 if one of the boundaries areabsorbing.
To ease the exposition we will assume that there are nonatural boundaries but all the results will continue to holdwhen a natural boundary is present after appropriate
modifications.
Note that if f ∈ H and vanishes at the absorbing
boundaries, thenPt f =∞
Xe−λn t (φn , f )φn .

n=0Thus, if g = Pt f , we can recover f from g via
f =

∞
X

eλn t (φn , g)φn .

n=0
Umut Çetin

Linear inverse problems and their regularisation

The above gives us a recipe to solve our inverse problem.
In order to determine the domain of Pt−1 first note thatg ∈ S ∩ H ∩ D(A∞ ) if g = Pt f for t > 0 whenever f ∈ H.

Umut Çetin

Linear inverse problems and their regularisation

The above gives us a recipe to solve our inverse problem.
In order to determine the domain of Pt−1 first note thatg ∈ S ∩ H ∩ D(A∞ ) if g = Pt f for t > 0 whenever f ∈ H.
Let D(Pt−1 ) be the class of functions, g, in S ∩ H ∩ D(A∞ )such that∞Xe2λn t (g, φn )2 < ∞.

n=0

Then, we may definePt−1 : g ∈ D(Pt−1 ) 7→∞X

eλn t (g, φn )φn .

n=0

It is worth to note that the sum in the above definition
converges absolutely and uniformly on the compact
squares of I × I.

Umut Çetin

Linear inverse problems and their regularisation

Note, however, that since λn → ∞, eλn t → ∞, indicatingthat Pt−1 is an unbounded operator.

Umut Çetin

Linear inverse problems and their regularisation

Note, however, that since λn → ∞, eλn t → ∞, indicatingthat Pt−1 is an unbounded operator.
Unboundedness of the operator in particular implies that
one cannot have a solution for every g and the domain of
the operator PT−1 is dense.
The last observation is both bad and good news: Even if
we cannot have a solution to the original problem, we canget arbitrarily close to a solution.

Umut Çetin

Linear inverse problems and their regularisation

Note, however, that since λn → ∞, eλn t → ∞, indicatingthat Pt−1 is an unbounded operator.
Unboundedness of the operator in particular implies that
one cannot have a solution for every g and the domain of
the operator PT−1 is dense.
The last observation is both bad and good news: Even if
we cannot have a solution to the original problem, we canget arbitrarily close to a solution.
Remark: If A were bounded, i.e. if X were a Markov chain,then Pt−1 would have the full domain being a bounded
operator.

Umut Çetin

Linear inverse problems and their regularisation

An example
The simplest example of functions in D(Pt−1 ) is g = p(u, z, ·) foru > t and z ∈ I. Indeed, sinceg=∞X

e−λn u φn (z)φn ,n=0

Umut Çetin

Linear inverse problems and their regularisation

An example
The simplest example of functions in D(Pt−1 ) is g = p(u, z, ·) foru > t and z ∈ I. Indeed, sinceg=∞X

e−λn u φn (z)φn ,

n=0
∞
X

e2λn t (g, φn )2 =n=0

Umut Çetin

Linear inverse problems and their regularisation

An example
The simplest example of functions in D(Pt−1 ) is g = p(u, z, ·) foru > t and z ∈ I. Indeed, sinceg=∞X

e−λn u φn (z)φn ,

n=0
∞
X
n=0

e2λn t (g, φn )2 =∞X

e−2λn (u−t) φ2n (z)n=0Umut Çetin

Linear inverse problems and their regularisation

An example
The simplest example of functions in D(Pt−1 ) is g = p(u, z, ·) foru > t and z ∈ I. Indeed, sinceg=∞X

e−λn u φn (z)φn ,

n=0
∞
X

e2λn t (g, φn )2 =

n=0

∞
X

e−2λn (u−t) φ2n (z) = p(2(u − t), z, z) < ∞.

n=0Moreover,Pt−1 g(x) =∞Xe−λn (u−t) φn (z)φn (x) = p(u − t, z, x),n=0

Umut Çetin

Linear inverse problems and their regularisation

An example
The simplest example of functions in D(Pt−1 ) is g = p(u, z, ·) foru > t and z ∈ I. Indeed, sinceg=∞X

e−λn u φn (z)φn ,

n=0
∞
X

e2λn t (g, φn )2 =

n=0

∞
X

e−2λn (u−t) φ2n (z) = p(2(u − t), z, z) < ∞.

n=0Moreover,Pt−1 g(x) =∞Xe−λn (u−t) φn (z)φn (x) = p(u − t, z, x),n=0

as one would also guess from the semi-group relationZp(u, z, x) =p(t, x, y )p(u − t, z, y )m(dy )Iin view of the symmetry of p.
Umut Çetin

Linear inverse problems and their regularisation

How to determine whether g ∈ D(Pt−1 )It turns out that there is a ‘simple’ criterion that will provide
the answer to our question.
Define the operator J α for α > 0 by
J α g(t, x) :=

Z ∞
0

√
J0 (2 ts)e−αs Ps g(x)ds,

where J0 is the Bessel function of the first kind of order 0. If
X is transient α can be taken equal 0.
One can show easily that t 7→ (J α g(t, ·), g) is convex anddecreasing to 0 as t → ∞.

Umut Çetin

Linear inverse problems and their regularisation

Coming back to our original question of whether g belongsto D(PT−1 ), the answer lies in the tail behaviour of(J α g(t, ·), g).
In fact, it is fairly easy to show that g ∈ D(PT−1 ) if and only if
Z ∞
0

Z ∞
√
√
J0 (2 ts)e−αs (Ps g, g)dsdt < ∞,I0 (2 2Tt)
0where I0 is the modified Bessel function of the first kind oforder 0.
We also have the inversion formula
Z ∞
√ Z ∞
√
PT−1 g = e−αTI0 (2 Tt)
J0 (2 ts)e−αs Ps gdsdt. (3)00Umut Çetin

Linear inverse problems and their regularisation

Alternative construction using a Picard iteration
For λ ≥ 0 consider the λ-potential operator:
U λ :=

Z ∞0e−λt Pt dt.

Then, J α g is the unique solution of the following Cauchy
problem:dj(t, ·) = −U α j(t, ·)dtj(0, ·) = U α g.

Umut Çetin

Linear inverse problems and their regularisation

Alternative construction using a Picard iteration
For λ ≥ 0 consider the λ-potential operator:
U λ :=

Z ∞0e−λt Pt dt.

Then, J α g is the unique solution of the following Cauchy
problem:dj(t, ·) = −U α j(t, ·)dtj(0, ·) = U α g.
Consequently, if we set
j0 (s, ·) = U α g,
jn+1 (s, ·) = U α g −Z t
0

U α jn (r , ·)dr ,(jn (·, ·))n≥0 converges in L2 ([0, t] × E, ds × m) to (Jsα g)s∈[0,t]
if t < α, where ds denotes the Lebesgue measure on [0, t].
Umut Çetin

Linear inverse problems and their regularisation

Regularisation
The most common method in practice is the Tikhonov
regularisation, which in our set up corresponds to the solutionof
Pt f + γf = g,
γ > 0, g ∈ L2 (m).
The above is a special case of the following:
Theorem 1
Suppose that φ : R+ 7→ R+ is continuous with
lim infx→∞ φ(x) > 0 and supx≥0 e−tx φ(x) < ∞. Then,(1 − γ)Pt f + γφ(−A)f = g,γ ∈ (0, 1),(4)

has a unique solution for any g ∈ L2 (m) and t > 0. Moreover,the solution has the property that
(1 − γ)f = arg min kPt h − gk2 +h∈L2 (m)Umut Çetin

γ(Pt φ(−A)h, h).
1−γ

(5)

Linear inverse problems and their regularisation

The assumption that lim infx→∞ φ(x) > 0 cannot be
dispensed easily if (4) is to have a solution for any giveng ∈ L2 (m). To wit take φ(x) = e−tx . Then (4) becomes
Pt f = g, which does not have a solution in general.
For each γ ∈ (0, 1) denote by fγ the solution of (4). Assume
further that g ∈ D(Pt−1 ). Thenlim kfγ − Pt−1 gk = 0.

γ→0Although looking abstract the above theorem furnishes us
with a plethora of concrete examples for regularising theinverse problem. Before giving some concrete examples
let us consider the following corollary.

Umut Çetin

Linear inverse problems and their regularisation

Mixing with jump processes
Suppose that K is a bounded positive operator such thatK = ψ(−A) for some bounded continuous function
ψ : R+ 7→ [0, 1] and γ ∈ (0, 1) is arbitrary. In an enlargement of
the probability space there exists a Markov process Y such thatY = ξX + (1 − ξ)J,
where ξ is a Bernoulli random variable with P(ξ = 1) = γ, J is ajump Markov process with generator
Bf = Kf − f ,f ∈ L2 (m),and ξ, J and X are mutually independent. Moreover,D(Qt−1 ) = L2 (m), where (Qt ) is the semigroup associated to Y .

Umut Çetin

Linear inverse problems and their regularisation

Examples

Fix T ∗ > 0 and let K = PT ∗ . Then, J is a Markov jumpprocess that remains constant between the jumps of a
Poisson process with unit parameter and moves between
the points of E according to the transition function PT ∗ or issentRto the cemetery state with probability
1 − E p(T ∗ , x, y )m(dy ).
In particular, when X is a Brownian motion, the process Y
is a Brownian motion with probability γ and a compound
Poisson process with normally distributed jumps withprobability 1 − γ .

Umut Çetin

Linear inverse problems and their regularisation

Examples

Fix T ∗ > 0 and let K = PT ∗ . Then, J is a Markov jumpprocess that remains constant between the jumps of a
Poisson process with unit parameter and moves between
the points of E according to the transition function PT ∗ or issentRto the cemetery state with probability
1 − E p(T ∗ , x, y )m(dy ).
In particular, when X is a Brownian motion, the process Y
is a Brownian motion with probability γ and a compound
Poisson process with normally distributed jumps withprobability 1 − γ .
One can also consider K = λU λ for some λ > 0.

Umut Çetin

Linear inverse problems and their regularisation

What is missing?
A comparison result:
Recall that Falkner (1983) has shown (under duality
assumption and another mild condition) for a generaltransient Markov process, X , with potential operator U that
if Uµ ≤ Uν for measures µ and ν, then one can find astopping time τ such that Xτ has law µ if ν is the
distribution of X0 .

Umut Çetin

Linear inverse problems and their regularisation

What is missing?
A comparison result:
Recall that Falkner (1983) has shown (under duality
assumption and another mild condition) for a generaltransient Markov process, X , with potential operator U that
if Uµ ≤ Uν for measures µ and ν, then one can find astopping time τ such that Xτ has law µ if ν is the
distribution of X0 .
A comparison result of this nature would be really useful.

Umut Çetin

Linear inverse problems and their regularisation

What is missing?
A comparison result:
Recall that Falkner (1983) has shown (under duality
assumption and another mild condition) for a generaltransient Markov process, X , with potential operator U that
if Uµ ≤ Uν for measures µ and ν, then one can find astopping time τ such that Xτ has law µ if ν is the
distribution of X0 .
A comparison result of this nature would be really useful.
Note that in order for g ∈ D(PT−1 ) it is necessary thatUg ≤ Uh for some h ∈ D(PT−1 ).

Umut Çetin

Linear inverse problems and their regularisation

What is missing?
A comparison result:
Recall that Falkner (1983) has shown (under duality
assumption and another mild condition) for a generaltransient Markov process, X , with potential operator U that
if Uµ ≤ Uν for measures µ and ν, then one can find astopping time τ such that Xτ has law µ if ν is the
distribution of X0 .
A comparison result of this nature would be really useful.
Note that in order for g ∈ D(PT−1 ) it is necessary thatUg ≤ Uh for some h ∈ D(PT−1 ).
However, this necessary condition is not sufficient: Let gbe the distribution of Xτ , where τ = inf{t ≥ T : |Xt | > a}and X is a standard Brownian motion killed as soon as itexits a finite interval. g cannot be in D(PT−1 ) since it has apoint mass.
Umut Çetin

Linear inverse problems and their regularisation